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To solve this problem, we need to maximize the amount of light that the Norman window can admit, which involves maximizing the area of the window subject to the given constraint on the perimeter.

### Step 1: Understand the geometry
The Norman window consists of:
- A rectangular part of width \( w \) and height \( h \).
- A semicircular part with radius \( r \), where the diameter of the semicircle is equal to the width of the rectangle, meaning \( w = 2r \).

### Step 2: Express the perimeter
The perimeter consists of:
- The perimeter of the semicircle: \( \text{Semicircular perimeter} = \frac{1}{2} \times 2\pi r = \pi r \).
- The perimeter of the rectangular sides (excluding the top, as it's already part of the semicircle): \( \text{Rectangular perimeter} = 2h + w \), and since \( w = 2r \), we can substitute this.

So, the total perimeter \( P \) is given by:
\[
P = 2h + 2r + \pi r
\]
We are given that the total perimeter is 23 feet:
\[
2h + 2r + \pi r = 23
\]
This is the constraint equation.

### Step 3: Express the area
The area \( A \) of the window is the sum of the area of the rectangle and the area of the semicircle:
\[
A = \text{Rectangular area} + \text{Semicircular area} = w \times h + \frac{1}{2}\pi r^2
\]
Substitute \( w = 2r \) into the equation:
\[
A = 2r \times h + \frac{1}{2} \pi r^2
\]

### Step 4: Solve the perimeter equation for \( h \)
From the perimeter equation:
\[
2h + (2 + \pi)r = 23
\]
Solve for \( h \):
\[
2h = 23 - (2 + \pi)r
\]
\[
h = \frac{23 - (2 + \pi)r}{2}
\]

### Step 5: Substitute \( h \) into the area equation
Substitute the expression for \( h \) into the area equation:
\[
A = 2r \times \frac{23 - (2 + \pi)r}{2} + \frac{1}{2}\pi r^2
\]
Simplify:
\[
A = r \times (23 - (2 + \pi)r) + \frac{1}{2}\pi r^2
\]
\[
A = 23r - (2 + \pi)r^2 + \frac{1}{2}\pi r^2
\]
Combine terms:
\[
A = 23r - \left( (2 + \pi) - \frac{1}{2} \pi \right) r^2
\]
\[
A = 23r - \left( 2 + \frac{\pi}{2} \right) r^2
\]

### Step 6: Maximize the area
To find the value of \( r \) that maximizes the area, we take the derivative of \( A \) with respect to \( r \) and set it equal to zero:
\[
\frac{dA}{dr} = 23 - 2 \times \left( 2 + \frac{\pi}{2} \right) r = 0
\]
Solve for \( r \):
\[
23 = 2 \times \left( 2 + \frac{\pi}{2} \right) r
\]
\[
r = \frac{23}{2 \times \left( 2 + \frac{\pi}{2} \right)}
\]
Simplify the denominator:
\[
r = \frac{23}{2 \times \left( 2 + 1.57 \right)} \approx \frac{23}{7.14}
\]
\[
r \approx 3.22 \, \text{ft}
\]

### Step 7: Find \( w \) and \( h \)
Since \( w = 2r \):
\[
w = 2 \times 3.22 \approx 6.44 \, \text{ft}
\]

Now, substitute \( r \) back into the perimeter equation to find \( h \):
\[
2h + 2 \times 3.22 + \pi \times 3.22 = 23
\]
\[
2h + 6.44 + 10.12 \approx 23
\]
\[
2h + 16.56 \approx 23
\]
\[
2h \approx 6.44
\]
\[
h \approx 3.22 \, \text{ft}
\]

### Final Answer:
- \( r \approx 3.22 \, \text{ft} \)
- \( w \approx 6.44 \, \text{ft} \)
- \( h \approx 3.22 \, \text{ft} \)

These dimensions maximize the amount of light the window can admit.

Do you need further clarification or details?

Here are 5 related questions:
1. How does the radius affect the perimeter and area of the window?
2. What would happen to the area if the perimeter constraint was increased?
3. How does the presence of a semicircle alter the optimization compared to a simple rectangle?
4. What are the units used for the area of the window?
5. Can we solve this problem using a different method, such as Lagrange multipliers?

**Tip:** For optimization problems involving geometry, it's often helpful to first express the constraints and objective function in terms of the same variables.

If the perimeter of the Norman window is 23 ft, find the dimensions of r and w of the window that allow the window to admit the most light.

Learn how to solve the problem of finding the dimensions of a Norman window that maximize the light it can admit, with a fixed perimeter of 23 feet. The solution uses key concepts in geometry and optimization.

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